# Show $e_1\wedge\cdots \wedge e_d$ is a non-zero vector in $\Lambda_d (V)$

Given $\{e_1,\cdots,e_d\}$ a basis of $V$, we have $e_1\wedge\cdots \wedge e_d$ is a non-zero vector in the exterior algebra $\Lambda_d (V)$.

Given a decomposable element which can be written as $v_1\wedge\cdots\wedge v_d$, it is the $0$ vector in $\Lambda_d (V)$ if $v_i = v_j$ for some $i,j$ between $1$ and $d$.

So in general, the finite sum of decomposable elements like above are also zero. How would I argue that $e_1\wedge\cdots \wedge e_d$ can not be written as the sum of such elements.

• It seems difficult to get to a proof in the way you suggest. It is easier to find a $d$-linear map which is non-zero under insertion of $e_1,\dots,e_d$ (essentially the determinant). – Andreas Cap Nov 27 '16 at 9:11
• More generally, the wedge product vanishes if and only if the $v_i$ are linearly dependent. – Ted Shifrin Nov 27 '16 at 16:31

Assuming you know some basic properties of determinants, here's one way you can do it. Define a map $D:V^d\to k$ (where $k$ is the scalar field) by taking a $d$-tuple of vectors to the determinant of the $d\times d$ matrix formed by taking those vectors as the columns (using the basis $\{e_1,\dots,e_d\}$). The determinant of a matrix is linear in each column of the matrix, and is $0$ if two columns are the same, so $D$ is an alternating multilinear map. By the universal property of $\bigwedge^d(V)$, there is a unique linear map $\tilde{D}:\bigwedge^d(V)\to k$ such that $\tilde{D}(v_1\wedge\dots\wedge v_d)=D(v_1,\dots,v_d)$ for any $v_1,\dots,v_d\in V$. In particular, $\tilde{D}(e_1\wedge\dots\wedge e_d)=D(e_1,\dots,e_d)$. But the matrix with columns $e_1,\dots,e_d$ is just the identity matrix, so $D(e_1,\dots,e_d)=1$. Thus $\tilde{D}(e_1\wedge\dots\wedge e_d)=1\neq 0$, so $e_1\wedge\dots\wedge e_d\neq 0$.